Definition. Let be an index set. Given a set , we define a -tuple of elements of
to be a function . If is an element of , we often denote the value of at by ranther than ; we call it the coordinate of . And we often denote the function itself by the symbol which is as close as we can come to a "tuple notation" for an
arbitary index set . We denote the
set of all -tuples of elements of
by
Definition. Let be an indexed family of sets; let
. The cartesian product of this indexed family, denoted
by is defined to be the set of all -tuples of elements
of such that for each . That is, it is the set of
all functions such that for each .
NOTE: The Cartesian product is defined as:
This is the set of all functions such that for every .
EXAMPLE:,
A function from to could be:
This function can be represented as the tuple:
The Cartesian product is the set of all such tuples:
Occasionally we denote the product simply by , and its
general element by , if
the index set is understood.
If all the sets are
equal to one set , then the
cartesian product is just the set of all -tuples of elements of . We sometimes use "tuple notation" for
the elements of , and
sometimes we use functional notation, depending on which is more
convenient.
Definition. Let be an indexed family of
topological spaces. Let us take as a basis for a topology on the product
space the collection of all sets of the form where is open
in , for each . The topology generated by
thsis is called the box topology
This collection satisfies the first condition for a basis because
is
itself a basis element.
And it satisfies the second condition because the intersection of any
two basis elements is another basis element.
Now we generalize the subbasis formulation of the
definition.
Let be the function assigning to each element of the product space
its coordinate,
it is called the projection mapping
associated with the index .
Definition. Let denote the collection
and let
denote the union of these collections, The topology generated by the
subbasis is called the
product topology. In this topology is
called a product space.
To compare these topologies, we consider the basis that generates
Intersection of Elements in
- Intersecting elements from the same set does not create anything new
because:
- However, new elements of arise when intersecting
elements from different sets.
Typical Element of
- A "typical element" of is constructed by:
- Choosing a finite set of distinct indices from .
- Letting be open sets
in .
- Taking the intersection:
Membership of Points in
- A point is in if:
- Its -th coordinate lies in
,
- Its th coordinate lies in
,
- And so on.
- There are no restrictions on for indices .
General Structure of
- The basis can be
written as: where:
- is the entire
space if ,
- Otherwise, is the chosen
open set .
is formed by
combining open sets from
various , restricting
coordinates specified by indices , and leaving other coordinates
unrestricted. This forms a basis for
the product topology.
All this is summarized in the following theorem:
Theorem 19.1(Comparison of the box and product
topologies).The box topology on
has as basis all sets of the form ,
where is open in for each .
The product topology on has
as basis all sets of the form ,
where is open in for each and equals except for finitely many
values of .
TWo things are immediately clear:
For finite products the
two topologies are precisely the same.
the box topology is in general finer
than the product topology
By the way, we prefer the product topology to the box topology. The
answer will appear as we continue our study of topology. We shall find
that a number of important theorems about finite products will also hold
for arbitary products if we use the product topology, but not if we use
the box topology. As a result, the product topology is extremely
important in mathematics. The box topology is not so important; we shall
use it primarily for constructing counterexamples, Therefore, we make
the following convention:
Whenever we consider the product , we
shall assume it is given the product topology unless we specifically
state otherwise.
Some of the theorems we proved for the product hold for the product no
matter which topology we use. We list them here.
Theorem 19.2. Suppose the topology in each space
is given by a basis
. The
collection of all sets of the form where for each , will serve as a basis for the
topology on .
The collection of all sets of the same form, where for
finitely many indices and
for all the
remaining indices, will serve as a basis for the product topology .
proof. First,
given , we find a basis such
that for
each . Hence we have
. And given , then we have , we can find a basis such that for each , due to is a basis for space . Thus, .
Above all, serve as a basis for the box topology on .
For the product topology, it is the same way to check that serve as a basis for the product topology
.
EXAMPLE 1 Consider euclidean -space . A basis for consists of all open intervals
in ; hence a basis for
the topology of
consists of all products of the form Since is a
finite product, the box and product topologies agree. Whenever we
consider , we will
assume that it is given this topology, unless we specifically state
otherwise.
Theorem 19.3. Let be a subspace of , for each . Then is a subspace of if both products are given
the box topology, or if both products are given the product
topology.
Theorem 19.4. If each space is a Hausdorff space, then is a Hausdorff space in
both the box and product topologies.
Theorem 19.5. Let be an indexed family of
spaces; let for each .
If is given either
the product or the box topology, then
proof.Let be a point of we show that
.
Let be a basis
element for either the box or product topology that contains . Since , we can
choose a point for each .
Then belongs to both
and . Since is arbitrary, it follows that belongs to the closure of .
Conversely, suppose
lies in the closure of , in either topology. We show that for any given index
, we have . Let be an arbitrary open set of containing . Since is open in in either topology, it
contains a point of
. Then belongs to . It follows that
.
Theorem 19.6. Let be given by the
equation where for each . Let
have the product topology. Then the function is continuous if and only if each
function is
continuous.
proof. Let be the projection of the
product onto its
factor. The function is
continuous, for if is
open in , the set is a subbasis
element for the product topology on . Now suppoose that is continuous. The function ; being
the composite of two continuous functions, it is continuous.
Conversely, suppose that each coordinate function is continuous. To prove that
is continuous, it suffices to
prove that the inverse image under of each subbasis element is open in
.
A typical subbasis element for the product topology on is
a set of the form , where is some index and is open in . Now because . Since
is continuous, this set is open in , as desired.
NOTE: The theorem fails if we use the box topology.
See the example below.
EXAMPLE 2 Consider , the countably infinite
product of with itself.
Recall that
where for each
. Let us define a function by
the equation
the th coordinate function of
is the function . Each of the coordinate
functions is continuous; therefore, the function is continuous if is given the product
topology. But is not continuous
if is given the
box topology. Consider, for example, the basis element
for the box topology. We assert that is not open in . If were open in , it would contain some
interval about
the point . This would mean that
, so
that, applying to both sides
of the inclusion,
The concept of continuous function is basic to much of mathematics.
Continuous functions on the real line appear in the first pages of any
calculus book, and continuous functions in the plane and in space follow
not far behind. More general kinds of continuous functions arise as one
goes further in mathematics. In this section, we shall formulate a
definition of continuity that will include all these as special cases,
and we shall study various properties of continuous functions. Many of
these properties are direct generalizations of things you learned about
continuous functions in calculus and analysis.
Continuity of a Function
Let and be a topological spaces. A function
is said to be
continuous if for each open subset of , the set is an open subset of .
Recall that is the set
of all points of for which ; it is empty if does not intersect the image set of .
Continuity of a function depends not only upon the function itself, but also on the topologies
specified for its domain and range. If we wish to emphasize this fact,
we can say that is
continuous relative to specific topologies on
and .
Let us note that if the topology of the range space is given by a basis , then to prove continuity of
it suffices to show that the
inverse image of every basis element is open. Then arbitrary open set
of can be written as a union of basis
elements Then so that is open
if each set is
open.
If the topology on is given by
a subbasis , to prove
continuity of it will even
suffice to show that the inverse image of each subbasis element is open.
The arbitrary basis element for
can be written as a finite
intersection of subbasis element; it follows from the equation that the inverse image of every basis element is open.
EXAMPLE 1 Let us consider a function like those
studied in analysis, a "real-valued function of a real variable," In analysis, one defines continuity of f via the " definition". As one
would expect, the
definition and ours are equivalent To prove
that our definition implies the definition, for instance, we proceed as follows.
Given in , and given , the interval is an open set of the range space . Therefore, is an open set
in the domain space .
Because contains the
point , it contains some basis
element about . We choose to be the smaller of the two
numbers and . Then if , the point must be in , so that , and , as
desired.
Conversely, given in , and given , , if , we have . Let , then , thus is a open set, because is a basis
element, as desired.
EXAMPLE 2 In calculus one considers the property of
continuity for many kinds of functions. For example, one sudies
functions of the following types: Each of them has a notion of continuity defined for it. Our
general definition of continuity includes all these as special cases,
this fact will be a consequence of general theorems we shall prove
concerning continuous functions on product spaces and on metric
spaces.
EXAMPLE 3 Let denote the set of real numbers
in its usual topology, and let denote the same set in the
lower limit topology. Let be the indentity function. for every real number . Then is not a continuous function, the
inverse image of the open set
of equals itself,
which is not open in . On
the other hand, the identity function is continuous, because the inverse image of is itself, which is open in .
Theorem 18.1. Let and be topological spaces; let . Then the following are
equivalent:
is continuous.
For every subset of , one has .
For every closed set of
, the set is closed in .
For each and each
neighborhood of , there is a neighborhood of such that .
If the condition in (4) holds for the point of , we say that is continuous at the point
.
proof. We show
that (1) (2) (3) (1) and that (1) (4) (1).
(1) (2). Assume that
is continuous. Let be a subset of . We show that if , then . Let be a neighborhood of . Then is an open set of containing ; it must intersect in some point . THen intersects in the point , so that , as
desired.
(2) (3). Let be closed in and let . We wish to prove that is closed in ; we show that . By elementary set theory, we
have .
Therefore, if , so that . Thus , so that . Thus , so that , as desired.
(3) (1). Let be an open set of . Set . Then Now is a closed set of
. Then is closed in by hypothesis, so that is open in , as desired.
(4) (1). Let be an open set of ; let be a point of . Then , so that by hypothesis there
is a neighborhood of such that . Then . If follows that
can be written as the
union of the open sets , so that
it is open.
(1) (4). Let and let be a neighborhood of . Then the set is a neighborhood of such that .
Homeomorphisms
Let and be topological spaces; let be a
bijection. If both the function and the inverse function are continuous, then is called a
homeomorphism.
The condition that be
continuous says that for each open set of , the inverse image of under the map is open in . But the inverse image of under the map is the same as the image of under the map . See Figure 18.1. So another way to
define a homeomorphism is to say that it is a bijective correspondence
such that is open if and only if is open.
This remark shows that a homeomorphism gives us a bijective correspondence not only between
and but between the collections of open
sets of and of . As a result, any property of that is entirely expressed in terms of
the topology of (that is, in
terms of the open sets of )
yields, via the correspondence ,
the corresponding property for the space . Such a property of is called a topological
property of
Remark: We canWe can think of homeomorphism as an
isomorphism in topology.
Now suppose that is
an injective continuous map, where and are topological spaces. Let be the image set , considerd as a subspace of ; then the function obtained by restricting
the range of is bijective.
If happens to be a
homeomorphism of with , we say that the map is a topological
imbedding, or simply an
imbedding, of in .
EXAMPLE 4 The function given by
is ahomeomorphism.
If we define by the equation then one can check easily that and for all real numbers and . It follows that is bijective and that , the continuity of and is a familiar result from calculus.
EXAMPLE 5 The function defined by is a homeomorphism. Its inverse is the function defined by
EXAMPLE 6 A bijective function can be continuous
without being a homeomorphism.
One such function is the identity map considered in Example 3.
Another is the following. Let
denote the unit circle. considered as asubspace of the plane , and let be the map defined by . The fact that is a bijective and contiuous follows
from familiar properties of the trigonometric functions. But the
function is note continuous.
The image under of the
open set of the domain, for instance, is not open in
, for the point lies in open set of such that . See Figure
18.4.
NOTE: The key that is not a homeomorphism is the is a open set for
EXAMPLE 7 Consider the function obtained from the function of the preceding example by expanding
the range. The map is an example
of a continuous injective map that is not an
imbedding.
Constructing Continuous
Functions
Theorem 18.2(Rules for constructing continuous
functions). Let , and
be topogical spaces.
(Constant function) If maps all of into the single point of , then is continuous.
(Inclusion) If is a subspace of , the inclusion function is continuous.
(Composites) If and are
continuous, then the map is continuous.
(Restricting the domain) If is continuous, and if is a subspace of , then the restricted function is continuous.
(Restricting or expanding the range) Let be continuous. If is a subspace of containing the image set , then the function obtained by restricting the
range of is continuous. If is a space having as a subspace, then the function obtained by expanding the
range of is continuous.
(Local formulation of continuity) The map be continuous if can be written as the union of open
sets such that is continuous for each
proof. (a) Let
for every in . Let be open in . The set equals or , depending on whether contains or not. In either case, it is
open.
(b) If is open in , then , which is open in
by definition of the subspace
topology.
(c) If is open in , then is open in and is open in . But by elementary set theory. (d) The function equals the composite of the
inclusion map and the map
, both of which are
continuous.
(e) Let is continuous. If
, we show
that the function
obtained from is continuous. Let
be open in . Then for some open set
of . We have by elementary set theory. Since is open, so is
To show is continuous if
has as a subspace, note that is the composite of the map and the inclusion map . (f) By hypothesis, we can
write as a union of open set
, such that is continuous for each
. Let be an open set in . Then so that is also
open in .
Theorem 18.3(The pasting lemma). Let , where and are closed in . Let and be
continuous. If for
every , then and combine to give a continuous function
, defined by setting
if , and if .
proof. Let be closed subset of . Now by elementary set theory. Since is continuous, is closed in and , therefore closed in . Similarly, is closed in and therefore closed in . Their union is thus closed in .
This is just a special case of the "local formulation of continuity"
rule given in preceding theorem.
EXAMPLE 8 Let us define a function by setting
Key Points:
Each "piece" ( and ) is continuous within its
domain.
The pieces overlap at ,
and both pieces evaluate to ,
ensuring continuity.
The function is defined
as:
Key Points:
At the overlapping point , the left piece evaluates to and the right piece to .
The pieces do not meet, creating a discontinuity at .
Another Example
The function (l(x)) is defined as:
Key Points:
The pieces do not overlap, making it a valid function.
However, is not
continuous:
There is a jump discontinuity at .
The inverse image of an open set is not open.
Theorem 18.4(Maps into products). Let be given by the
equation The is continuous if
and only the functions are continuous.
The maps and are called the coordinate
functions of .
proof. Let and be projections
onto the first and second factors, respectively. These maps are
continuous. For and , and these sets are open if and are open. Note that for each , If the function is
continuous, then and are composites of continuous
functions and therefore continuous.
Conversely, suppose that and
are continuous. We show that
for each basis element
for the topology of , its
inverse image is
open and .
Therefore, Since both of the sets and are open, so is their
intersection.
NOTE: There is no useful cntenon for the continuity
of a map
whose domain is a product space.
One might conjecture that I is continuous if it is continuous "in
each vanable separately," but this conjecture not
true.(See the problem below)
Let be defined by the equation:
Show that is continuous in
each variable separately.
Compute the function defined by .
Show that is not continuous.
Parametrized Curves in Calculus
A parametrized curve in the plane is defined as
a continuous map , often expressed in the form: where and are continuous functions of .
The continuity of relies
on the continuity of and .
Vector Fields in the Plane
A vector field in the plane is expressed as:
and are scalar functions that represent the
components of the vector field.
is continuous if
both and are continuous functions, which
is equivalent to saying that is a continuous map of into .
Constructing Continuous Functions via Limits
Another method to construct continuous functions is through
limits of infinite sequences of continuous
functions:
If a sequence of continuous functions converges
uniformly to a limit function , then is also continuous.
This result is called the Uniform Limit Theorem,
which is a fundamental theorem in analysis.
面对有关矩阵的题目，我们也要注意将矩阵看作一个线性变换，例如一个可逆矩阵可看作一个同构映射；一个上三角矩阵，可看作不变子空间分解；一个分块对角矩阵，可看作线性空间的直和分解，有根子空间直和分解，循环子空间直和分解，还有不变子空间直和分解，三者分别对应Jordan
form、frobenius form 和 直接对角化。
A subset of a topological
space is said to be
closed if the set is open.
EXAMPLE 1 The subset of is closed because its
complement is open. Similarly, is closed, because its
complement is open.
These facts justify our use of the terms "closed interval" and "closed
ray" The subset of is neither open nor
closed.
EXAMPLE 2 In the plane , the set is closed, because its complement is the union of the two sets
each of which is a product of open sets of and is, therefore, open in
EXAMPLE 3 In the finite complement
topology on a set , the
closed sets consist of itself and
all finite subsets of
finite complement topology
The finite complement topology (or cofinite topology) on a set is defined by specifying the open sets
in the topology:
1.The empty set and
the set itself are open.
2.Any subset is
open if and only if its complement is finite.
This topology is particularly useful when studying properties of
convergence and compactness, as it ensures that every sequence in has a convergent subsequence.
EXAMPLE 4 In the discrete topology on the set , every set is open, it
follows that every set is closed as well.
EXAMPLE 5 Consider the following subset of the real
line: in the subspace topology. In this space, the set is open, since it
is the intersection of the open set of with . Similarly, is open as a subset of ; it is even open as a subset of . Since and are complements in of each other, we conclude that both
and are closed as subsets of .
Let's talk about the properties of the closed sets.
Theorem 17.1. Let be a topological space. Then the
following condition hold:
and are closed.
Arbitrary intersections of closed sets are closed.
Finite unions of closed sets are closed.
proof. (1) and are closed because they are the
complements of the open sets and
, respectively.
(2) Given a collection of closed sets , we apply DeMorgan's law, Since the sets are open by definition, the right side of this
equation represents an arbitrary union of open sets, and is thus open.
Therefore, is
closed.
(3) Similarly, if is closed for
, consider the
equation The set on the right side of this equation is a finite
intersection of open sets and is therefore open, Hence is closed.
We should notice the conceptions of "closed set" is relative.
Now when dealing with subspaces, one needs to be careful in using the
term "closed set." If is a
subspace of , we say that a set
is closed in if is a subset of and if is closed in the subspace
topology of (that is, if
— is open in )
We have the following theorem:
Theorem 17.2. Let be a subspace of . Then a set is closed in if and only if it equals the
intersection of a closed set of
with .
proof. Assume
that , where is closed in . (See Figure 17.1) Then is open in , so that is open in , by the definition of the subspace
topology. But .
Hence is open in , so that is closed in . Conversely, assume that is closed in . (See Figure 17.1) Then is open in , so that by definition it equals the
intersection of an open set of
with . The set is closed in , and , so that
equals the intersection of a closed set of with , as desired.
Figure 17.1
Theorem 17.3. Let be a subspace of . If is closed in and is closed in then is closed in .
proof. If is closed in , then equals the intersection of a closed set
of with , by the theorem 17.2, so we have . And is also closed in , thus is closed in .
Closure and Interior of a
Set
Given a subset of a
topological space , the
interior of is
defined as the union of all open sets contained in .
The closure of is defined as the intersection of all
closed sets containing
The interior of is denoted by
Int and the closure of is denoted by . Obviously Int is an open set and is a closed set; furthermore,
If is open , ; while if is closed, .
We shall not make much use of the intenor of a set, but the closure
of a set will be quite important.
NOTE:
When dealing with a topological space and a subspace , one needs to exercise care in taking
closures of sets If is a subset
of , the closure of in and the closure of in will in general be different In such a
situation.
We reserve the notation to
stand for the closure of in . The closure of in can be expressed in terms of .
Theorem 17.4. Let be a subspace of , let be a subset of , let denote the closure of in . Then the closure of in equals .
proof. Let denote the closure of in . The set is closed in , so is closed in in
Theorem 17.2. Since
contains , and since by definition
equals the intersection of all
closed subsets of containing
, we must have
On the other hand, we know that
is closed in . Hence by Theorem
17.2, for some set
closed in . Then is a closed set containing ; because is the intersection of all such
closed sets, we conclude that . Then .
Since the collection of all closed sets in , like the collection of all open sets,
is usually much too big to work with. Another way of
describing the closure of a set, useful because it involves only
a basis for the topology of , is given in the following
theorem.
Theorem 17.5. Let be a subset of the topological space
.
Then if and
only if every open set containing
intersects .
Supposing the topology of
is given by a basis, then if and only if every basis element containing intersects .
proof. Consider
the statement in (a). It is a statement of the form . Let us transform
each implication to its contrapositive, thereby obtaining the logically
equivalent statement (not ) (not ). Written out, it is the following.
In this form, our theorem is easy to prove. If is not in , the set is an open set containing
that does not intersect , as desired. Conversely, if there
exists an open set containing
which does not intersect , then is a closed set containing . By definition of the closure , the set must contain , therefore, cannot be in . Statement (b) follows readily if
every open set containing
intersects , so does every basis
element containing , because is an open set. Conversely, if every
basis element containing
intersects , so does every open
set containing , bacause contains a basis element that contains
.
Mathematicians often use some special terminology here. They shorten
the statement " is an open set
containing " to the phrase
Using this terminology, one can write the first
half of the preceding theorem as follows:
If is a subset of the
topological space , then if and only if every
neighborhood of intersects .
EXAMPLE 6 Let
be the real line . If
, then , for every neighbrhood of
intersects , while every point outside has a neighborhood disjoint from
Similar arguments apply to the
following subsets of .
If , then . If , then . If is
the set of rational numbers, then . If is
the set of positive integers, then . If
is the set of positive
reals, then the closure of is the set .
EXAMPLE 7 Consider the subspace of the real line . The set is subset of , its closure in is the set , and its closure in
is the set
Limit Point
There is yet another way of describing the closure of a set, a way
that involves the important concept of limit point, which we consider
now.
If is a subset of the
topological space and if is a point of , we say that is a limit
point (or "cluster point," or "point of accumulation") of
if every neighborhood of
intersects in some point other than itself. Said differently,
is a limit point of if it belongs to the closure of
. The point
may lie in or not; for this definition it does not
matter.
EXAMPLE 8 Consider the real line . If , then the point is a limit point of and so is the point . In fact, every point of the
interval is a limit point of
, but no other point of is a limit point of .
If , then is
the only limit point of . Every
other point of has a neighborhood that either
does not intersect at all, or it
intersects only in the point
itself. If , then the limit
points of are the points of the
interval .
If is the set of
rational numbers, every point of is a limit point of . If is the set of positive
integers, no point of is
a limit point of . If
is the set of positive
reals, then every point of is a limit point of .
Theorem 17.6. Let be a subset of the topological space
, let be the set of all limit points of
. Then
proof. If in , every neighborhood of intersects (in a point different from x).
Therefore, by Theorem 17.5,
belong to . Hence . Since by
definition , it
follows that .
To demonstrate the reverse inclusion, we let be a point of and show that . If happens to lie in , it is trivial that ; suppose that does not lie in . Since , we know that every neighborhood of intersects ; because , the set must
intersect in a point different
from . Then , so that , as desired.
Corollary 17.7 A subset of a topological space is
closed if and only if it contains all its limit points.
proof. The set
is closed , and the latter holds
.
Hausdorff Spaces
One's expenence with open and closed sets and limit points in the
real line and the plane can be misleading when one considers more
general topological spaces.
For example, in the spaces and , each one-point
set is closed.
This fact is easily proved; every point different from has a neighborhood not intersecting
, so that is its own closure.
But this fact is not true for arbitrary topological
spaces. Consider the topology on the three-point set indicated in Figure 17.3. In
this space, the one-point set is not closed, for its complement () is not open.
Figure 17.3
And, in topological space , a
subset of can have many different limit
points.
Similarly, one's experience with the properties of convergent
sequences in and can be misleading when one
deals with more general topological spaces. In an arbitrary topological
space, one says that a sequence of points of the space to the point of provided that, corresponding to each
neighborhood of , there is a positive integer such that for all . In and , a sequence cannot
converge to more than one point, but in an arbitrary space, it
can. In the space indicated in Figure 17.3, for example, the
sequence defined by setting
for all converges not only to the
point , but also to the
point and to the point !
Topologies in which one-point sets are not closed, or in which
sequences can converge to more than one point, are considered by
many mathematicians to be somewhat strange. They are not really
very interesting, for they seldom occur in other branches of
mathematics.
Therefore, one often imposes an additional condition that
will rule out examples like this one, bringing the class of spaces under
consideration closer to those to which one's geometric intuition
applies. The condition was suggested by the mathematician
Felix Hausdorff, so mathematicians have come to call it
by his name.
Definition. A topological space is called a Hausdorff
space if for each pair of distinct points of , there exists neighborhoods , and of and , respectively, that are disjoint.
Theorem 17.8. Every finite point set in a Hausdorff
space is closed.
proof. It
suffices to show that every one-point set is closed. If is
a point of different from , then and have disjoint neighborhoods and , respectively. Since does not intersect , the point cannot belong to the closure of set
. As a result, the closure
of the set is itself, so that is is
closed.
The condition that finite point sets be closed is in fact
weaker than the Hausdorff condition.
For example, the real line in the finite
complement topology is note a Hausdorff space, but it is a
space in which finite point sets are closed.
The condition that finite point sets be closed has been given a name
of its own: it is called the axiom
Theorem 17.9. Let be a space satisfying the axiom; let be a subset of . Then the point is a limit point of if and only if every neighborhood of
contains infinitely many points
of .
proof. If every
neighborhood of intersects in infinitely many points, it certainly
intersects in some point other
than itself, so that is a limit point of .
Conversely, suppose is a limit
point of , and suppose some
neighborhood of intersects in only finitely many points. Then
also intersects in finitely many points; let
be the
points of . The
set is
an open set of , since the finite
point set is
closed; then is a neighborhood of x that intersects the set not at all. This contradicts
the assumption that is a limit
point of .
One reason for our lack of interest in the axiom is the fact that many
of the interesting theorems of topology require not just that axiom, but
the full strength of the Hausdorff axiom. Furthermore, most of
the spaces that are important to mathematicians are Hausdorff spaces.
The following two theorems give some substance to these remarks.
Theorem 17.10. If is a Hausdorff space, then a sequence
of points of converges to at most
one point of .
proof. Suppose
that is a sequence of points of
that converges to . If , let and be disjoint neighborhoods of and , respectively. Since contains for all but finitely many values of
, the set cannot. Therefore, cannot converge to .
If the sequence of points of
the Hausdorff space converges to
the point of , we often write , and we say that is the limit
of the sequnce .
Theorem 17.11.
(1). Every simply ordered set is a Hausdorff space in the order
topology.
(2). The product of two Hausdorff spaces is a Hausdorff space.
(3). A subspace of a Hausdorff space is a Hausdorff space.
proof.
(1). Given two elements of
the simply orderd set , and . We can find an interval
be the neighborhood of , and be the neighborhood of such that . Thus,
, is a Hausdorff space.
(2). Suppose that and are the two Hausdorff spaces, and is the product topology. Given
, then
. We can find a neighborhood of , and of such that , we also can
find a neighborhood of , and of such that . Hence , and . So is a Hausdorff space.
(3). Suppose is a Hausdorff
space, . Obviously,
is also a Hausdorff space.
If is a simply ordered set,
there is a standard topology for ,
defined using the order relation. It is called the order
topology; in this section, we consider it and study some
of its properties.
Suppose that is set having
simple order relation . Given
elements and of such that , there are four subsets of that are called the intervals
determined by and . They are the following:
The notation used here is familiar to you already in the case where X
is the real line, but these are intervals in
an arbitrary ordered set. A set of the first type is called an
open interval in X, a set of the last type is
called a closed interval in X, and sets of the
second and third types are called half-open
intervals
Definition. Let be a set with a simple order relation;
assume has more than one element.
Let be the collection
of all sets of the following types:
All open intervals in
.
All intervals of the form , where is the
smallest element (if any) of .
All intervals of the form , where is the
largest element (if any) of .
The collection is a
basis for a topology on , which is
called the order topology.
if has no smallest element,
there are no sets of type (2), and if X has no largest element, there
are no sets of type (3)
Note: The elements of the order topology are not the
intervals, we should notice the definition of the
interval!!!
We need to check that satisfies the requirements
for a basis.
First, note that every element
of lies in at least one element
of : The smallest
element (if any) lies in all sets of type (2), the largest element (if
any) lies in all sets of type (3), and every other element lies in a set
of type (1).
Second, note that the intersection of any two sets of the preceding
types is again a set of one of these types, or is empty.
EXAMPLE 1 The standard topology on , as defined in the preceding
section is just the order topology derived from the usual order on
EXAMPLE 2 Consider the set in the
dictionary order, we shall denote the general element of by , to avoid difficulty with
notation. The set has neither a largest nor a smallest
element, so the order topology on , so the order topology on has as basis
the collection of all open intervals of the form for , and for , and . These two types of intervals are
indicated below.The subcollection consisting of only intervals of the
second type is also a basis for the order topology on , it is easy
to check, just think a moment.
EXAMPLE 3 The positive integers form an orederd set with a
smallest element. The order topology on is discrete topology, for
every one-point set is open If , then the one-point set is a basis element;
and if , the one-point set is a basis element.
EXAMPLE 4 The set in the dictionary order is another
example of an ordered set with a smallest element. Denoting by and by , we can represent by The order topology on
is not the discrete topology. Most one-point sets are open, but there is
an exception --- the one-point set .Any open set containing b1 must contain a basis element
about (by definition), and any
basis element containing
contains points of the
sequence.
Rays
Definition. If is an ordered set, and is an element of , there are four subsets of that are called the
rays determined by They are the following: Sets of the first two types are called open
rays, and sets of the last two types are called
closed rays
The use of the term "open" suggests that open rays in X are open sets
in the order topology. And so they are.
Consider, for example, the ray . If has a
largest element , then equals the basis element
. If has no largest element, then equals the union of all basis
elements of the form , for
. In either case, is open. A similar argument
applies to the ray .
The open rays, in fact, form a subbasis for the order topology on
, as we now show.
Because the open rays are open in the order topology, the topology
they generate is contained in the order topology.
On the other hand, every basis element for the order topology equals a
finite intersection of open rays; the interval equals the intersection of and , while and , if they exist, are themselves
open rays. Hence the topology generated by the open rays contains the
order topology.
The Product Topology on X × Y
If and are topological spaces, there is a
standard way of defining a topology on the cartesian product .
Definition. Let and be topological spaces. The
product topology on is the topology having as a
basis the collection of
all sets of the form ,
where is an open subset of and is an open subset of .
Let's check that is
a basis.
The first condition is trivial, since is itself a basis element. The second condition is
almost as easy, since the intersection of any two basis elements and is another basis element.
For and the latter set is a basis element because and are open in and , respectively. See the figure below, it
can help you understand it intutively.
Note: The collection is not a topology on . The union of the two
rectangles pictured in the figure above, for instance, is not a product
of two sets, so it is cannot belong to ; however, it is open in .
We may ask what's the basis of the topology of
Theorem 15.1. If is a basis for the topology
of and is a basis for the topology
of , then the collection is a basis for the topology of .
proof. We apply
Lemma 13.2. Given an open set of
and a point of , by definition of the product topology
there is a basis element
such that . Because and
are bases for and , respectively, we can choose an element
such that , and an element such that . Then . Thus
the collection meets
the criterion of Lemma 13.2, so is a basis for .
EXAMPLE 1 We have a standard topology on the order topology. The
product of this topology with itself is called the standard
topology on . It has as basis the collection of
all products of open sets of , but the theorem just proved
tells us that the much smaller collection of all products of open intervals in
will also serve as a
basis for the topology of . Each such set can be
pictured as the interior of a rectangle in
It is sometimes useful to express the product topology in
terms of a subbasis. To do this, we first define certain
functions called projections.
Definition. Let be defined by the equation let
be defined by the equation The maps and are called the projections of onto its first and second
factors, respectively.
We use word "onto" because
and are surjective.
If is an open subset of , then the set is precisely the set , which is open in . Similarly, if is open in , then , which is
also open in . The
intersection of these two sets is the set , as indicated in Figure below.
This fact leads to the following theorem:
Theorem 15.2. The collection is a subbasis for the product topology on .
proof. Let denote the product topology
on . let be the topology
generated by . Because
every element of
belongs to , so do the
arbitrary unions of finite intersections of elements of . Thus . On
the other hand, every basis element for the topology of is a finite intersection of
elements of , since
Therefore, , so that ,
hence , as desired.
The subspace Topology
Definition. Let be a topological space with topology
. If is a subset of , the collection is a topology on ,
called the subspace topology. With this
topology, is called a
subspace of ; its open sets consists of all
intersections of open sets of
with .
It is easy to see that is a topology.
It contains and because where and are elements of .
The fact that it is closed under finite intersections and arbitrary
unions follows from the equations
Lemma 16.1. If is a basis for the topology
of then the collection is a basis for the subspace topology on .
proof. Given
open in and given , we can choose an element such that . Then . It
follows from Lemma 13.2 that is a basis for the subspace
topology on .
When dealing with a space and
a subspace , one needs to be
careful when one uses the term "open set". Does one mean an
element of the topology of or an
element of the topology of ?
We make the following definition If is a subspace of , we say that a set U is open in
(or open relative to
) if it belongs to the topology of
; this implies in particular that
it is a subset of . We say that
U is open in if
it belongs to the topology of
There is a special situation in which every set open in is also open in .
Lemma 16.2. Let be a subspace of . If is open in and is open in , then is open in .
proof. Since
is open in , for some set open
. Since and are both open in , so is .
Now let us explore the relation between the subspace topology and the
order and product topologies.
Theorem 16.3. If is a subspace of and is a subspace of , then the product topology on is the same as the topology
inherits as a subspace of
.
proof.The set
is the general basis
element for , where is open in and is open in . Therefore, is the
general basis element for the subspace topology on . Now Since and are the general open sets fot
the subspace topologies on the
and , respectively, the set is the general
basis element for the product topology on .
Thus, we know that the bases for the subspace topology on and for the product topology on
are the same. Hence the
topologies are the same.
Now let be an ordered set in
the order topology, and let be a
subset of . The order relation on
, when restricted to , makes into an ordered set.
However, the resulting order topology on need not be the same as the topology
that inherits as a subspace of
. We give one example
where the subspace and order topologies on agree, and two examples where they do
not.
EXAMPLE 1 Consider the subset of the real line , in the subspace topology. The
subspace topology has as basis all sets of the form , where in an open interval in . Such a set is of one the
following types. By definition, each of these sets is open in . But sets of the second and third types
are not open in the larger space .
Note that these sets form a basis for the order topology on
Y. Thus, we see that in the case of the set , its subspace topology
(as a subspace of ) and
its order topology are the same.
EXAMPLE 2 Let
be the subset of
. In the subspace
topology on the the one-point set
is open,
because it is the intersection of the open set with . But in the order topology on , the set is not open. Any basis element for the
order topology on that contains
is of the form for some , such a
set necessarily contains points of less than .
EXAMPLE 3 Let . The dictionary order on is just the restriction to of the dictionary order on the
plane .
However, the dictionary order topology on is not the same as the subspace
topology on obtained
from the dictionary order topology on ..
For example, the set is open in in the subspace topology, but
not in the order topology, because we choose , any basis element for the order topology on
that contains is of the form
such a set is necessarily not be contained in
The set in the
dictionary order topology will be called the order
square, and denoted by
Convex
Given an ordered set , let us
say that a subset of is convex in
if for each pair of points of , the entire interval (a, b) of points
of lies in . Note that intervals and rays in are convex in .
Theorem 16.4 Let be an orderd set in the order topology;
let be a subset of that is convex in . Then the order topology on is the same as the topology inherits as a subspace of .
proof. Consider
the ray in . What is its intersection with ? If , then this is an open ray of the orderd set . If , then is either
a lower boud on or an upper bound
on , since is convex. In the former case, the set
equals all of
; in the latter case, is is
empty.
A similar work remark show that the intersection of the ray with is either an open ray of , or itself, or empty. Since the sets and form a subbasis for
the subspace topology on , and
since each is open in the order topology, the order topology contains
the subspace topology.
To prove the reverse, note that any open ray of Y equals the
intersection of an open ray of
with , so it is open in the
subspace topology on . Since
the open rays of are a
subbasis for the order topology on , this topology is contained in
the subspace topology, hence they are the same.
Comments: In this proof, we need to be familiar with
the definition of rays and its properties (such that it
is the subbasis for the order topology)
To avoid ambiguity, let us agree that whenever is an ordered set in the order topology
and is a subset of , we shall assume that is given the subspace topology unless
we specifically state otherwise. If is convex in , this is the same as the order topology
on , otherwise, it may not be.
Definition. If is any subset of , the exterior measure of
is where the infimum is taken over all countable coverings by
closed cubes. In general, the exterior measure is always non-negative
butt could be infinite, hence we have
NOTE:
It is important to note that it would not suffice to allow finite
sumsin the definition of
One can, however, replace the coverings by cubes, with covering
by rectangles; or with coverings by balls. That the former alternative
yields the same exterior measure is quite direct.
EXAMPLE 1. The exterior measure of a point is zero.
The exterior measure of the empty set is also zero.
This is clear that a point is a cube with volume zero, and which
covers itself, so does the empty set.
EXAMPLE 2. The exterior measure of a closed cube is
equal to its volume.
Suppose is a closed cube in
. Since covers itself, we have .Therefore, it suffices to
prove the reverse inequality.
We consider an abitraty covering by cubes, and by the definition of
the exterior measure, it suffices to prove that For a fixed we choose for each an
open cube which contains , and such that . From the
open covering of the compact set , we may select a finite subcovering,
which after possibly renumbering the rectangles, we may write as . Taking
the closure of the cubes , we
have . Consequently, Since is
arbitrary, we find the inequality before we need to prove holds; thus
, as desired.
EXAMPLE 3. If
is an open cube, the result still holds.
Since is covered by its
closure , and , we immediately see
that , and obviously
by the definition
of exterior measure.
EXAMPLE 4. The exterior measure of a rectangle is equal to its volume.
Arguing as in Example 2, we see that . To obtain the reverse inequality, consider a grid
in formed by cubes of
sides length . Then, if
consists of the
(finite) collection of all cubes entirely contained in , and the (finite) collection
of all cubes that intersect the complement of , we first note that . Also, a simple argument yields Moreover, there are cubes in , and these cubes have
volume , so that . Hence and letting tend to
infinity yields , as
desired.
EXAMPLE 5. The exterior measure of is infinite.
This follows from the fact that any covering of is also a covering of any
cube , hence
. Since
can have arbitrarily large
volume, we must have
EXAMPLE 6. The Cantor set has exterior measure .
From the construction of , we know that , where
each is a disjoint
union of closed intervals,
each of length .
Consequently, for all , hence .
Observation 1 (Monotonicity) If , then .
This follows that any coverings of by a countable collection of cubes is
also a covering of .
In particular, monotonicity implies that every bounded subset of
has finite exterior
measure.
Observation 2 (Countable sub-additivity) If , then
.
First, we may assume that each , for otherwise the inequality clearly holds. For
any , the definition
of exterior measure yields for each a covering
by closed cubes with Then, is a covering of by closed cubes, and therefore Since this holds true for every , the second observation is
proved.
Observation 3 If , then , where the infimum is taken over all open
sets containing .
By monotonicity, it is clear that the inequality holds. For the
reverse inequality, let and choose cubes
such that , with Let denote an
open cube containing , and such
taht . Then is open, and by Observation 2
Hence , as was shown.
Observation 4 If , and , then
By Observation 2, we already know that , so it is
suffices to prove the reverse inequality. To this end, we first select
such that . Next, we
choose a covering by closed cubes, with . We may, after subdividing the cubes , assume that each has a diameter less than . In this case, each can intersect at most one of the two
sets or . If we denote by and the sets of those indices for which intersects and , respectively, then is empty, and we have Therefore
Observation 5 If a set is the countable union of
almost disjoint cubes, then
Let denote a cube
strictly contained in such that
, where is a arbitrary but fixed. Then,
for every , the cubes are disjoint, hence at a finite distance from one
another, and repeated applications of Observation 4 imply Since , we conclude that for every integer , When the tend to
infinity, we deduce for every , hence , as desired.
NOTE:
This last property shows that if a set can be decomposed into almost
disjoint cubes, its exterior measure equals the sum of the volumes of
the cubes. Moreover, this also yields a proof that the sum is
independent of the decomposition
Measureable sets and
the Lebesgue measure
Definition. A subset of is Lebesgue
measurable or simply measureable, if for any
there exists an
open set with and
If is measurable, we define
its Lebesgue measure (or measure)
by Clearly, the Lebesgue measure inherits all the features
contained in Observations 1 - 5 of the exterior measure.
Let's study the properties of the Lebesgue measure.
Property 1 Every open set in is measurable.
Property 2 If , then is measurable.
In particular, if is a subset of
a set of exterior measure , then
is measurable.
By Observation 3 of the exterior measure, for every there exists an open set
with and . Since
, monotonicity implies , as
desired.
Property 3 A countable union of measurable sets is
measurable.
Suppose , where each is
measurable. Given ,
we may choose for each an open
set with and . Then the union is open, , nd , so monotonicity and sub-additivity of
the exterior measure imply
Property 4Closed sets are
measurable
First, we observe that it suffices to prove that compact sets
are measurable. Indeed, any closed set can be written as the union of compact
sets, say , where denotes the
closed ball of radius centered at
the origin; then Property 3 applies.
So suppose is compact (so that in
particular ), and
let . Since is
closed, the difference is open, and by Theorem 1.4 we may write this difference as a
countable union of almost disjoint cubes FOr a fixed , the finite
union is
compact; therefore
(we isolate this little fact in a lemma below). Since ,
Observation 1, 4, and 5 of the exterior measure imply Hence , and
this also holds in the limit as
tends to infinity. Invoking the sub-additivity property of the exterior
measure finally yields as desired.
Theorem 1.4. Every open subset of , , can be written as a countable
union of almost disjoint closed cubes.
Lemma 3.1. If
is closed, is compact, and these
sets are disjoint, then .
proof. Since
is closed, for each point , there exists so that . Since
covers , and is compact, we may find a subcover,
which we denote by . If
we let , then we must have . Indeed, if
and , then for some we have , and by
construction . Therefore and the lemma is proved.
Compact set: A bounded set is
compact if it is also closed. Compact sets
enjoy the Heine-Borel covering property.
Property 5 The complement of a measurable set is
measurable.
If E is measurable, then for every positive integer we may choose an open set with and .
The complement is
closed, hence is measurable, which implies that the union is also measurable by Property 3. Now we
simply note that ,
and such that for all .
Therefore, , and
is measurable by Property
2. Therefore is measurable
since it is the union of two measurable sets, namely and .
Property 6 A countable intersection of measurable
sets is measurable.
This follows from Properties 3 and 5, since
Warning:
we have proved that the collection of measurable sets is closed under
countable unions and intersections. We emphasize, however, that the
operations of uncountable unions or intersections are
not permissible when dealing with measurable sets!
Theorem 3.2. If are disjoint measurable sets, and , then
proof. First, we
assume further that each is
bounded. Then, for each , by
applying the definition of measurability to , we can choose a closed subset
or with . For each fixed , the set are compact and
disjoint, so that . Since , we must
have Letting tend to
infinity, since was
arbitrary we find that Since the reverse inequality always holds(by sub-aditivity in
Observation 2), this concludes the proof when each is bounded.
In the general case, we select any sequence of cubes that increases
to , in the sense
that for all
and . We then let and for . If we
define measurable sets by . then The union above is disjoint and every is bounded. Moreover , and
this union is also disjoint. Putting these facts together, and using
what has already been proved, we obtain as claimed.
We make two definitions to state succinctly some further
consequences.
If is a countable collection of subsets of that increases to in the sense that for all , and , then we write
Similarly, if
decreases to in the sense that
for all k,
and ,
we write .
Corollary 3.3 Suppose are measurable subsets
of .
If , then
.
If and for some k, then
proof. For the
first part, let , , and in general for . By their construction, the sets
are measurable, disjoint, and
.
Hence and since we get the desired limit.
>For the second part, we may clearly assume that . Let for each , so that
is a disjoint union of measurable sets. As a result, we find that Hence, since , we see that , and the proof is
complete.
NOTE: The reader should note that the second
conclusion may fail without the assumption that for some . This is shown by the simple example
when , for all .
What follows provides an important geometric and analytic
insight into the nature of measurable sets, in terms of their
relation to open and closed sets.
an arbitrary measurable set can be well approximated by the
open sets that contain it, and alternatively, by the closed sets it
contains.
Theorem 3.4 Suppose is a measurable subset of . Then, for every :
There exists an open set with and .
There exists a closed set
with and .
If is finite, there
exists a compact set with and .
If is finite, there
exists a finite union of closed cubes such that The notation
stands for the symmetric difference between the sets and , defined by , which
consists of those points that belong to only one of the two sets or .
proof. Part(i) is
just the definition of the measurability.
For the second part, we know that is measurable, so there exists an
open set with and . If
we let , then is closed, , and
. Hence
as
desired.
Comments: If we meet the closed set, we can consider
its complement, because its complement is open set.
For (iii), we first pick a closed set so that and . For each , we let denote the ball centered at the
origin of radius and define
compact sets . Then
is a sequence of measurable
that decreases to , and since
, we conclude that
all large one has .
For (iv), choose a family of closed cubes so that Since ,
the series converges and exists such that . If , then
Invariance properties
of Lebesgue measure
Translation-invariance: If is a measurable set and , then the set is alse
measurable, and .
To prove the measurability of under the assumption that is measurable, we note if is open, , and , then
is open, , and
Relative dilation-invariance: Suppose , and denote by the set . We can then
assert that is measurable
whenever is, and .
Reflection-invariant: Whenever is measurable, so is and .
σ-algebras and Borel sets
Definition. A - algebra of sets is a
collection of subsets of that is closed under
countable unions, countable intersections, and complements.
NOTE: We should compare the definition of - algebra with the topology. In
the definition of the topology, we only need to guarantee that
finite intersections is closed, but the union
can be arbitrary, instead of countable. This is a difference of
the two important definition.
The collection of all subsets of is of course a - algebra.
Another - algebra, which
plays a vital role in analysis, is the Borel - algebra in , denoted by , which by
definition is the smallest -
algebra that contains all open sets. Elements of this - algebra are called
Borel sets.
We should discuss the existence and uniqueness of the Borel -algebra.
The definition of the Borel -algebra will be meaningful once we
have defined the term “smallest,” and shown that such a -algebra exists and is unique. The
term "smallest" means that if is any -algebra that contains all open
sets , then
necessarily . Since we observe that any intersection
(not necessarily countable) of -algebras is again a -algebra, we may define as the
intersection of all -algebras
that contain the open sets. This shows the existence
and uniqueness of the Borel -algebra.
Since open sets are measurable, we conclude that the Borel -algebra is contained in the -algebra of measurable sets.
Naturally, we may ask if this inclusion is strict.
do there exist Lebesgue measurable sets which are not Borel sets? The
answer is “yes.”
Starting with the open and closed sets, which are the simplest Borel
sets. Next in order would come countable intersections of open
sets; such sets are called sets. Alternatively,
one could consider their complements, the countable union of
closed sets, called the sets.
Corollary 3.5 A subset of is measurable
if and only if differs
from a by a set of
measure zero.
if and only if differs
from an by a set of
measure zero.
proof. Clearly
is measurable whenever it
satisfies either (i) or (ii), since the , , and sets of measure zero are
measurable.
Conversely, if is measurable,
then for each integer we may
select an open set
that contains , and such that
. Then is a that contains , and for all . Therefore for all ; hence has exterior measure zero, and is
therefore measurable.
For the second implication, we simpy apply part (ii) of Theorem 3.4 with
, and take
the union of the resulting closed sets.
Construction of a
non-measurable set
Are all subsets of measurable? In this
section, we answer this question when by constructing a subset of which is not measurable. This
justifies the conclusion that a satisfactory theory of measure cannot
encompass all subsets of
The construction of a non-measurable set uses the axiom of choice, and
rests on a simple equivalence relation among real numbers in .
We write whenever is rational, and note that this is an
equivalence relation since the follwing properties hold:
for every
if , then
if and , then .
Two equivalence classed either are disjoint or coincide, and is the disjoint union of all
equivalence classed, which we write as Now we construct the set by choosing exactly one
element from each , and setting .
Theorem 3.6 The set is not measurable.
proof. The proof
is by contradiction, so we assume that is measurable. Let be an
enumeration of all the rationals in , and consider the translates We claim that the sets are disjoint, and To see why these sets are disjoint, suppose that the
intersection is non-empty. Then there exist rationals
and and with ;
hence Consequently and is rational; hence , which
contradicts the fact that contains only one
representative of each equivalence class.
The second inclusion is straightforward since each is contained in by construction. Finally, if , then for some , and therefore for some . Hence , and the first inclusion holds. Now we may
conclude the proof of the theorem. If were measurable, then so
would be for all , and since the union is
disjoint, the inclusions in (4) yield Since is a
tanslate of , we must
have for all .
Consequently, This is the desired contradiction, since neither nor is possible.
Axiom of choice
That the construction of the set is possible is based on the
following general proposition.
Suppose is a set and is a collection of
non-empty subsets of . (The
indexing set of 's is not
assumed to be countable.) Then there is a function (a "choice
function") such that for all .
In this general form this assertion is known as the axiom of
choice.
The initial realization of the importance of this axiom was in its
use to prove a famous assertion of Cantor, the well-ordering
principle. This proposition (sometimes referred to as
"transfinite induction") can be formulated as follows.
A set is linearly
oredered if there is a binary relation such that:
(a) x for all .
If are distinct, then
either or (but not both).
If and , then .
We say that a set can be
well-orederd if can be linearly oredered in such a way
that every non-empty subset has a smallest element in that ordering (that is, an element
such that for any other ).
EXAMPLE. A simple example of a well-ordered set is
, the positive
integers with their usual ordering. The fact that is well-ordered is an
essential part of the usual (finite) induction principle. More
generally, the well-ordering principle states:
Any set can be
well-ordered.
It is in fact nearly obvious that the well-ordering principle
implies the axiom of choice: if we well-order , we can choose to be the smallest element in
, and in this way we have
constructed the required choice function. It is also true, but not as
easy to show, that the converse implication holds, namely that the axiom
of choice implies the well-ordering principle.
The fact that the axiom of choice and the well-ordering principle are
equivalent is a consequence of the following considerations.
One begins by defining a partial ordering on a set to be a binary relation on the set that satisfies:
x for all .
If and , then .
If and , then .
If in addition or whenever , then is a linear ordering of .
The axiom of choice and the well-ordering principle are then
logically equivalent to the Hausdorff maximal principle:
Every non-empty partially ordered set has a (non-empty) maximal
linearly ordered subset.
In other words, if is patially
ordered by , then contains a non-empty subset which is linearly ordered by and such that if is contained in a set also linearly ordered by , then .
An application of the Hausdorff maximal principle to the collection
of all well-orderings of subsets of E implies the well-ordering
principle for E. However, the proof that the axiom of choice implies the
Hausdorff maximal principle is more complicated.
Measurable functions
With the notion of measurable sets in hand, we now turn our attention
to the objects that lie at the heart of integration theory: measurable
functions.
The starting point is the notion of a characteristic
function of a set , which
is defined by The next step is to pass to the functions that are the
building blocks of integration theory. For the Riemann integral it is in
effect the class of step functions, with each given as
a finite sum where each is a
rectangle, and the are
constants.
However, for the Lebesgue integral we need a more general notion, as
we shall see in the next chapter. A simple function is
a finite sum where each is a
measurable set of finite measure, and the are constants.
Definition and basic
properties
We begin by considering only real-valued functions on , which we allow to take on
the infinite values and
, so that belongs to the extended real numbers
We shall say that is
finite-valued if for all . In the theory that follows, and the
many applications of it, we shall almost always find ourselves in
situations where a function takes on infinite values on at most
a set of measure zero.
A function defined on a
measurable subset of is measurable, if for all
, the set is measurable. To sumplify our notation, we shall often denote
the set
simply by whenever no
confusion is possible.
